Consider the following half reactions: IrCl63 + 3e- Ir 1 6Cl2 o = 0.77 V PtCl42-
Question:
IrCl63– + 3e- → Ir 1 6Cl2 ϐo = 0.77 V
PtCl42- + 2e- → Pt 1 4Cl2 ϐo = 0.73 V
PdCl42– + 2e– → Pd 1 4Cl2 ϐo = 0.62 V
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